How to Remove Leading Zeroes from a String

CInt("000000000000512")

See conversion functions:

Use Clng if you expect to have large numbers, as already pointed out in a comment:

Clng("000000004000512")

otherwise you'll have an overflow, as variant's subtype int is 16 bit in vbscript

This will work even with a crazy long number

Function RemoveLeadingZeroes(ByVal str)
Dim tempStr
tempStr = str
While Left(tempStr,1) = "0" AND tempStr <> ""
    tempStr = Right(tempStr,Len(tempStr)-1)
Wend
RemoveLeadingZeroes = tempStr
End Function

strNewFileName = RemoveLeadingZeroes("0009283479283749823749872392384")

I've used this technique before:

• replace the zeros with spaces
• left trim
• replace the spaces with zeros

Replace(LTrim(Replace(str, "0", " ")), " ", "0")

Note, this doesn't work if str has meaningful spaces in it.

Use the Absolute Value of the number.

Var = ABS(Var)

Function TrimLeadingZeros(value)
    TrimLeadingZeros = value    
    while left(TrimLeadingZeros, 1) = "0" and TrimLeadingZeros <> "0"
        TrimLeadingZeros = mid(TrimLeadingZeros, 2)
    wend
End Function

or

Function TrimLeadingZeros(value)
    dim i
    i = 1
    while i < len(value) and mid(value,i,1) = "0"
        i = i + 1
    wend
    
    TrimLeadingZeros = mid(value, i)
End Function

Using regex:

Regex.Replace("000000000000512", "^0+", "") ' returns "512"
Regex.Replace("0000040000000", "^0+", "") ' returns "40000000"

In case your string includes digits and characters, use a Do While statement:

string = "00000456ABC"

Do While Left(string, 1) = "0"
     string = Right(string, (Len(string)-1))
Loop

Function TrimLZ(str)  
   If Left(str, 1) = "0" Then
      TrimLZ = TrimLZ(Mid(str, 2, Len(str)))
   Else
      TrimLZ = str
   End If
End Function